For solving: https://sudokupad.app/wzq81314ut

1. Branching thermometers suggest coloring.
1a. For example, R5C2 is a thermometer bulb, so whatever digit it contains, it cannot repeat anywhere on the rest of the thermometer. So in box 7, it cannot appear in R7C1, nor in column 3, and it is already in column 2. That leaves either R8C1 or R8C2. In box 1, since it already appears in column 1 and column 2, it must be in column 3.
1b. R6C2 similarly cannot appear again on the thermometer, nor again in column 2, so it also in either R8C1 or R8C2. In box 1, it is also in column 3.
1c. R4C2 is above the bulb, rather than below, so we look at box 1. It cannot repeat on the same thermometer, eliminating column 1. It is already in column 2. That leaves column 3. That means that column 2 in box 4 contains the same digits as column 3 in box 1, which forces whatever is in R4C2 into R7C1, so the same three digits are in column 1 in box 7.
1d. This indicates "roping." Whatever is in column 1 in box 1 cannot be in the same column in box 4, and we already know that column 2 in box 4 doesn't repeat in column 1 of box 1, so it repeats in column 3 of box 4, which means it also repeats in column 2 of box.
1e. Finally, whatever is in column 2 of box 1 repeats in column 1 of box 4 and column 3 of box 7. If we color R123C1 red, then we also color R456C3 and R789C2 red. R123C2, R456C1, and R789C3 can all be blue, leaving R123C3, R456C2, and R789C1 to be green.

2. While the other two large thermometers are similar, they are different enough that roping isn't forced as it is in the first three columns. Still, there are some similar patterns.
2a. R5C5 cannot repeat in column 5, nor on the thermometer, so in box 8 it appears in either R8C6 or R9C6.
2b. R6C5 is forced into the other of either R8C6 or R9C6.
2c. Those two cells are two of the three cells in column 4 of box 2, but the third is unknown.
2d. This is enough for something like partial roping: The three cells in box 2 column 6 cannot repeat in column 6, and also cannot repeat in R5C5 or R6C5, but that still leaves four cells in box 5 in which those three cells repeat: R4C4, R5C4, R6C4, and R4C5. Either there is unforced roping, or only two of the digits "rope" while the third slips into R4C5. It is still helpful to color, but multiple colors will be needed to allow for multiple possibilities.
2e. The same is true for box 2 column 5, which might repeat entirely in box 8 column 4, or it might repeat two cells in box 8 column 4, and the third in R7C6.

3. Another different thermometer shape, another pattern of coloring.
3a. R5C8 cannot repeat in column 8, nor on the thermometer, so it's in R8C9 or R9C9 in box 9. It's also in column 7 in box 3.
3b. Looking up, R4C8 also appears in column 7 in box 3, but it can be in any of the three cells in column 9 in box 9.
3c. R6C5 isn't on a thermometer, so nothing keeps it from repeating in either column 7 or 9 in boxes 3 and 9.
3d. We can still narrow things down somewhat. In box 3, the three cells in column 9 are on a thermometer, but can they repeat on the thermometer in box 9? Consider: if the three digits in box 3 column 9 repeated in box 9 column 7, they would have to then repeat in box 6 column 8, but we already know that two of those three cells are unavailable. At most, one digit may repeat in R6C5 (or not), but the other two (or all three) must be in box 9 column 8.
3e. The same is true for box 3 column 8, which may contribute at most one digit to R7C9, and therefore contributes either two or three digits to box 9 column 7. We can now have coloring for every cell.

4. Let's think about digits. In the first three columns, column 1 in box 1 is on a thermometer and the minimum possible value for any of the three cells is 3 because of the distance from the thermometer bulb. Column 3 in box 7 is also on a thermometer and the minimum possible value for any of the three cells is also 3, but it is a different three cells, for a total of six cells with a maximum value of 9 and a minimum value of 3, with one degree of freedom. That means the cells in box 4 column 2, which are green if colored as suggested in step 1, must include 1 and 2, along with another digit. Now that 1 is on the thermometer, there is only one cell into which it can go: R5C2 is 1.

5. Now 2 must be in either R4C5 or R6C5. If it is in R4C5, it repeats in R7C1 (see step 1c), and that's too far away from the thermometer bulb, so R6C5 is 2, and R8C1 and R9C1 are a 1-2 pair. Both 1 and 2 are in box 1 column 3, along with a third unknown digit that is also in R4C2 and R7C1.

6. With 1 in R5C2, where can 1 go in row 7? Six of the nine cells are on thermometer stems,  R7C2 is in a column that already has a 1, and R7C5 is even, so R7C8 is 1. That forces 1 in box 6 to be R4C9, since every other cell is either too far along a thermometer, or even, or shares a row or column with an existing 1. Together those force 1 in box 5 into R6C4. That means 1 in box 8 is in either R8C5 or R9C5, and in box 2 in either R2C6 or R3C6. It also resolves the coloring in R6C1 to match column 3 in box 2, but leaves the possibility of roping open. The coloring in the columns 7, 8, and 9 is also partially resolved, enough to know roping isn't happening there. The 1 in R4C9 and the 1 in R7C8 force a 1 into either R1C7 or R3C7, which means it matches with two of the three cells in box 9 column 9.

7. Back to row 7 now for 2: six cells in row 7 are too far from a thermometer bulb, column 2 already has a 2, and now column 8 is taken, so R7C5 is 2. We also know that 3 in box 7 is in row 7 using the same logic.

8. In fact, since roping tells us R7C2 is in column 1 in box 1, it can't contains a 3; all three of those (red) cells are too far along the thermometer. So 3 is in either R7C1 or R7C3. Back in step 3 we worked out that the three cells in box 3 column 9 repeat in three of four cells in column 8, but now one of those cells is a 1, which cannot appear in box 3 column 9. Now we know that R123C9 definitely repeat in column 8 in row 6, row 8, and row 9, and none of those digits can be lower than 4. Neither can any of the box 9 cells on a thermometer stem. That forces R8C9 and R9C9 to be a 2-3 pair, and one of those cells is repeated from R5C8, so R5C8 is also either 2 or 3. R5C5 cannot be 1 or 2, and if it is 5 it doesn't leave enough digits to fill all four stem cells in box 8, so R5C5 is either 3 or 4.

9. In row 9, there are now only three cells which can contain. 1, 2, or 3: in box 7, the 3 must be up in row 7, and R9C3 is too far along a thermometer, so R9C1 is 12, but that's it. In box 9, we just worked out R9C8 must be one of the digits from box 3 that is too far along a thermometer, while R9C7 is itself too far along a thermometer, leaving only R9C9, which is 23. We still need one more from box 8. In box 8, 2 is spoken for, and R9C6 is even, while R9C4 is too far along a thermometer, so R9C5 can be 1 or 3, and that's it. So that's a rotating 1-2-3 triple in columns 1, 5, and 9.

10. We can combine that with the question of where 1, 2, 3, and 9 go in row 8, or we can pencil in thermometer candidates and see that we have a 4-5-6-7-8 quintuple in that row. Either way, R8C5 and R8C6 can only be 1, 3, or 9. But R8C6 is repeated from either R5C5 or R6C5, neither of which can be 1 or 9, so it is 3, which means R5C5 is 3, and those 3s force R5C8 and R8C9 to be 2. R9C5 is now 1, R8C5 is 9, R9C9 is 3, R9C1 is 2, and R8C1 is 1. If we penciled in thermometer stems, the newly placed 9 narrows down those pencil marks.

11. There are a number of different ways to answer the question of where 4 goes in column 5, but let's consider this: There are now four thermometer cells in box 8 which must all be lower than 9, so together they are 5, 6, 7, and 8, with the 5 in row 7. Whichever one it is, R6C5 is 4. So is R9C6.

12. The thermometer that heads up from R5C5 can now be filled in with 5, 6, 7, and 8.

13. With 6 and 8 placed in box 2, the two remaining even cells must be a 2-4 pair. Since column 6 already has a 4, R1C4 is 4 and R1C6 is 2. Now 2 in box 5 can only be in one place: R4C4.

14. Let's pencil in all thermometer candidates if we haven't already. Even with that done, this next one might be tough to spot! Consider that 9 in column 9 has only two places it can go. (All other cells are either filled-in with 1, 2, or 3, or are even.) 9 in row 7 also has only two places it can go. (All other cells are either mid-stem on a thermometer, or mirror a cell which is mid-stem on a thermometer, or are filled-in with 1 or 2, or share a box with a placed 9.) R7C9 is part of both sets. To resolve which it is, consider what happens if R1C9 is 9. That would eliminate 9 from R7C9, but it would also push the 9 in box 1 into column 2, eliminating 9 from R7C2. Now there's no place for 9 in row 7 at all, so R1C9 is not 9, so R7C9 is 9.

15. That placement has a ripple effect, and there are many different directions to go, but here's one: Now we know that R7C9 is not the cell that mirrors R4C8, since 9 can't appear mid-thermometer, which resolves some of the coloring in box 9 and means that R9C9 is mirrored into R4C8, making it 3. That means the minimum value for R6C7 is now 5, which means the top of that thermometer must be either 8 or 9, but since we just placed 9 in box 9, R9C7 is 8, R8C7 is 7, R7C7 is 6, and R6C6 is 5. Since all of the other even digits are placed in box 9, R8C8 is 4, making R9C8 5. We now know the contents of column 7 in box 3, which must be 1, 2, and 3 in some order. One of those must be even, so R2C7 is 2. The only remaining digits in column 7 are 4 and 9, but putting 4 on the tip of even a two-cell thermometer requires the bulb to be lower than 4, which it cannot be, so R4C7 is 9 and R5C7 is 4. In box 9, we filled in the two cells that must appear in box 3 column 9, or another way to spot that is to ask where 4 and 5 are in column 9, or yet another way is to see that we've placed both 4 and 5 in column 7 and column 8, so they must be placed in column 9 in box 3. However we get there, R3C9 is 4 and R2C9 is 5.

16. Continuing to work out the effects of placing that 9, since we used up the 6-7-8 run in column 7, we can't repeat it in column 4, so that run starting at R7C4 is 5-6-7, and R7C6 is 8. That means the even digit in R6C6 is 6, which combines with a recently-place 9 on row 4 to place 7 in R4C6, 9 in R5C6, and 8 in R5C4. Now the even digit in R5C9 is 6, R1C9 is 7, R6C9 is 8, R6C8 is 7, R3C8 is 8, R2C8 is 9 via thermometer, and R1C8 is 6.

17. Shifting left, R2C6 is 1 so R3C6 is 5, R2C4 is 3 so R3C4 is 9, and R1C1 can no longer be 6, 7, or 8, so it is 9, while R3C1 can no longer be 4 or 5 (and hasn't been able to be 6 for some time), so it is 7, and R2C1 falls between 7 and 9 and is therefore 8. So many places to go next, but let's visit R3C3 which is even and not 4 or 6 or 8 so it is 2. We've known from the beginning that box 1 column 3 would need to have a 1 and a 2 in it, and we've placed the 2, and R2C6 means 1 is taken for that row, so R1C3 is 1, which resolves the 1 and 3 in box 3.

18. Speaking of 3, we can finally the question of what exactly is our third "green" cell along with 1 and 2 in boxes 1, 4, and 7. The answer is: not 3! In fact, there are so many digits placed in row 6 that the only ones we still need are 3 and 9, and there's already a 9 in column 1, so R6C1 is 3. That forces a 3 into R7C3, and combined with the other placed digits in row 7, forces a 4 into R7C1 and therefore also R4C2 and therefore also R2C3.

19. Each of the first three rows now has eight placed digits, so it should be easy to place the missing 3, 5, and 6 in box 1. Roping reduces the candidates for the remaining cells so that the order of each is forced by the already-placed digits. And that's the puzzle!